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a_sh-v [17]
3 years ago
13

A ______ is a push or pull that one object exerts on another.

Physics
2 answers:
Veronika [31]3 years ago
8 0

Answer:

A. Force

Explanation:

The answer is force

ICE Princess25 [194]3 years ago
5 0

Answer:

The answer is A

Explanation:

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What is the correct formula for power? A. Power = work / time B. Power = work * time C. Power = force * distance D. Power = work
steposvetlana [31]

Answer:

A. Power = Work / Time

Explanation:

Power is the amount of work done over time, or rather the rate of work, which is given by the unit of watts (W). Since work is defined by Force * Displacement, we can also say Power = Force * Displacement / Time.

3 0
3 years ago
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Define the difference between the rigid body problem and a single particle problem.
agasfer [191]
. In single particle problem whole mass is concentrated at a single point so it has a single displacement, single velocity and single acceleration. while, in rigid body mass is distributed
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The change in the wind patterns on the earth that causes an increase in ocean temperatures is
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3 years ago
1. A racing car with the driver weighs 1825 lb. Find the kinetic energy in ft*lb when traveling with a speed of 100 mi/hr.
Svetlanka [38]

Answer:

1. 610,000 lb ft

2. 490 J

Explanation:

1. First, convert mi/hr to ft/s:

100 mi/hr × (5280 ft / mi) × (1 hr / 3600 s) = 146.67 ft/s

Now find the kinetic energy:

KE = ½ mv²

KE = ½ (1825 lb / 32.2 ft/s²) (146.67 ft/s)²

KE = 610,000 lb ft

2. KE = ½ mv²

KE = ½ (5 kg) (14 m/s)²

KE = 490 J

6 0
3 years ago
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
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